Q 11-03-095JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
When vector $\vec A=2\hat i+3\hat j+2\hat k$ is subtracted from vector $\vec B$, it gives a vector equal to $2\hat j$. Then the magnitude of vector $\vec B$ will be
Answer: (D) $\sqrt{33}$
$\vec B=\vec A+2\hat j=2\hat i+5\hat j+2\hat k$, so $|\vec B|=\sqrt{4+25+4}=\sqrt{33}$.
Solution by Sreeraj P, M.Sc Physics